/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 40 When we want \(95 \%\) confidenc... [FREE SOLUTION] | 91Ó°ÊÓ

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When we want \(95 \%\) confidence and use the conservative estimate of \(p=0.5,\) we can use the simple formula \(n=1 /(M E)^{2}\) to estimate the sample size needed for a given margin of error ME. In Exercises 6.40 to 6.43, use this formula to determine the sample size needed for the given margin of error. A margin of error of 0.01

Short Answer

Expert verified
The required sample size for a margin of error of 0.01 is 10000.

Step by step solution

01

Formula Recognisation

The given formula to determine the sample size for the desired margin of error is \(n=1 /(M E)^{2}\), where ME is the margin of error.
02

Substitution of Values

Substitute the provided margin error ME = 0.01 into the formula. So the sample size \(n = 1 / (0.01)^{2}\).
03

Calculation

On computing \(1 / (0.01)^{2}\), we obtain 10000.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Margin of Error
When conducting a survey or a study, researchers aim to measure the views or characteristics of a whole population based on a sample from that population. However, the results from the sample are unlikely to match the entire population exactly, which is where the concept of the margin of error comes into play.

The margin of error is a statistic expressing the amount of random sampling error in a survey's results. It represents how much the results could vary if the survey were conducted multiple times. The margin of error is commonly presented as a plus or minus value (e.g., ±3 percent) and indicates the precision of the estimate. A smaller margin of error requires a larger sample size and suggests more confidence in the consistency of the survey results. Understanding the margin of error helps individuals interpret survey results accurately and gauge the level of certainty in the findings.
Confidence Level
The confidence level is a measure of the reliability of the results obtained from a statistical sample. It reflects how often the true population parameter (such as the mean or proportion) would be captured by the interval estimate if the sampling process were repeated numerous times. Typically expressed as a percentage, the confidence level represents the degree of certainty that the sample statistic will fall within a certain range.

For example, a 95% confidence level means that if you were to draw 100 different random samples and compute a confidence interval for each sample, approximately 95 of these intervals would be expected to contain the true population parameter. The confidence level is essential in determining the sample size: higher confidence levels, such as 99% or 95%, often require larger sample sizes to ensure that the margin of error is acceptably low.
Sample Size Formula
To accurately estimate a particular statistic of a population, such as a proportion or a mean, researchers use a sample size formula to determine how large their sample should be to achieve a desired margin of error at a certain confidence level. A fundamental formula used for this purpose, especially when estimating a proportion and assuming a 50% response distribution (the most conservative estimate), is \( n = \frac{1}{(ME)^2} \), where \( n \) is the sample size and \( ME \) is the margin of error.

Using this formula helps researchers to decide on the number of responses or observations needed in their study for reliable and valid results. Given a smaller margin of error and a high confidence level, the required sample size will increase. This is crucial, as underestimating the sample size can lead to inaccurate results, while overestimating can unnecessarily increase the cost and duration of the study.

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Most popular questions from this chapter

In Exercises 6.28 to 6.31, use StatKey or other technology to generate a bootstrap distribution of sample proportions and find the standard error for that distribution. Compare the result to the standard error given by the Central Limit Theorem, using the sample proportion as an estimate of the population proportion \(p\). Proportion of peanuts in mixed nuts, with \(n=100\) and \(\hat{p}=0.52\)

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We examine the effect of different inputs on determining the sample size needed to obtain a specific margin of error when finding a confidence interval for a proportion. Find the sample size needed to give a margin of error to estimate a proportion within \(\pm 3 \%\) with \(99 \%\) confidence. With \(95 \%\) confidence. With \(90 \%\) confidence. (Assume no prior knowledge about the population proportion \(p\).) Comment on the relationship between the sample size and the confidence level desired.

In Exercises 6.188 to 6.191 , use the t-distribution to find a confidence interval for a difference in means \(\mu_{1}-\mu_{2}\) given the relevant sample results. Give the best estimate for \(\mu_{1}-\mu_{2},\) the margin of error, and the confidence interval. Assume the results come from random samples from populations that are approximately normally distributed. A \(95 \%\) confidence interval for \(\mu_{1}-\mu_{2}\) using the sample results \(\bar{x}_{1}=75.2, s_{1}=10.7, n_{1}=30\) and \(\bar{x}_{2}=69.0, s_{2}=8.3, n_{2}=20 .\)

Use a t-distribution to answer the question. Assume the sample is a random sample from a distribution that is reason ably normally distributed and we are doing inference for a sample mean. Find the area in a t-distribution above 1.5 if the sample has size \(n=8\).

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